Does there exists a finite measure $\mu$ on the Lebesgue $\sigma$-algebra on $\mathbb{R}$ for which $\mu<Does there exists a finite measure $\mu$ on the Lebesgue $\sigma$-algebra on $\mathbb{R}$ for which $\mu<<m$ and $m<<\mu$ where m is the Lebesgue measure
I can see from this post that
for positive function $f\in L^1(\mathbb{R},\mathrm dm(x))$ and for $A$ a Lebesgue measurable set, the measure $\mu$ defined as bellow should work.
$$
\mu(A)=\int\limits_A f(x)\;\mathrm d m(x)
$$
But I would like some more clarification.
for the finiteness, I can say $\mu(R)=\int\limits_R f(x)\;\mathrm d m(x)\leq\int\limits_R |f(x)|\;\mathrm d m(x)<\infty$ as $f\in L^1(R,dm)$
But to prove the absolute continuity in both ways:
$
\begin{align}
\mu(A)=0 &\Rightarrow \int\limits_A f(x)\;\mathrm d m(x)=0\\
&\text{ But why this should imply } m(A)=0?
\end{align}$
And for
$\begin{align}
m(A)=0
\end{align}$
is it correct to say that all simple functions $\phi=\sum\limits_{k=1}^{n} a_{k}\chi_{A_k}$ you have
$\int\phi dm=\sum a_km(A_k)=0$ for all the subsets $A_k$ in $A$. And as $\int fdm$ is the suprimum of $\int\phi$ where $\phi\leq f$ we have $\int f=0$.
Thus
$\begin{align}
m(A)=0&\Rightarrow\int\limits_{A}fdm=0\\
&\Rightarrow \mu(A)=0
\end{align}$
Appreciate your help
 A: How about $$\mu(A) = \int_A \frac{1}{1+x^2} \, dm(x)?$$
Clearly $m(A) = 0 \implies \mu(A) = 0$. On the other hand for any $k \ge 1$ you may define $$E_k = \displaystyle \left\{ x \in \mathbf R \mid \frac{1}{1+x^2} \ge \frac 1 k \right\}$$
and kindly observe that
$$\mu(A) \ge \mu(A \cap E_k) = \int_{A \cap E_k} \frac{1}{1+x^2} \, dm(x) \ge \frac 1k m(A \cap E_k).$$
If $\mu(A) = 0$ then of course  $m(A \cap E_k) = 0$ for any $k$. As defined $\cup E_k = \mathbf R$ so that
$$m(A) \le \sum_k m(A \cap E_k) = 0$$ whenever $\mu(A) = 0$.
There was nothing too special about $\dfrac{1}{1+x^2}$ here. You could replace it with your favorite positive integrable function.
A: For any strictly positive function $f\in L_1(\lambda)$, where $\lambda$ is Lebesgue’s, the measure
$$ \mu_f= f\,d\lambda$$
satisfies the desired conditions. Indeed, by construction $\mu\ll\lambda$. On the other hand, $\lambda =\frac{1}{f}\,d\mu_f$, and so $\lambda\ll\mu_f$.
Here are some examples that appear in many applications:

*

*The Gaussian measure: $$ \phi(dx)=\frac{1}{\sqrt{2\pi}}e^{-\tfrac{x^2}{2}}\,dx$$

*The Cauchy measure: $$\nu(dx)=\frac{1}{\pi}\frac{dx}{1+x^2}$$

*The logistic measure: $$\mu(dx)=\frac{e^{-x}}{(1+e^{-x})^2}\,dx$$
